Re: If line L in the xy-coordinate plane has a positive slope,
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17 May 2021, 02:07
OFFICIAL EXPLANATION:
If line L in the xy-coordinate plane has a positive slope, what is the x-intercept of L ?
(1) There are different points (a, b) and (c, d) on line L such that ad = bc.
(2) There are constants m and n such that the points (m, n) and (–m, –n) are both on line L.
STATEMENT (1) SUFFICIENT: The easiest way to deal with this statement is to pick smart numbers, and see what the x-intercept is in each case. Plug in sets of numbers a, b, c, and d such that ad = bc. In every case, the x-intercept of the resulting line will be 0; plug in enough points to become convinced of this pattern.
For instance, let a = 6, b = 4, c = 3, d = 2. Then the line passes through the points (6, 4) and (3, 2). By finding the equation of the line we can determine that the x-intercept of the line is (0, 0).
Try another set of numbers: a = –10, b = –5, c = –2, d = –1. In this case, the line passes through (–10, –5) and (–2, –1). Again, by finding the equation of the line, we can determine that the x-intercept of the line is (0, 0).
Continue plugging until you are convinced that the pattern won’t have any exceptions. The x-intercept is always 0; the statement is sufficient.
Algebra:
Solve for one of the variables in the equation ad = bc, say d:
\(d = \frac{bc}{a}\)
Therefore, line L passes through the points \(( a, b) and (c, \frac{bc}{a} )\)
To write an equation of the line, find its slope: (y2-y1)/(x2-x1)=
\([(\frac{bc}{a} -b)] / [(c - a)]\)
Multiply the top and bottom of this fraction by a to eliminate the smaller denominator, giving \( \frac{[(bc - ba)]}{[a(c-a)]} = \frac{[b(c-a)]}{[a(c-a)]} = \frac{b}{a}\)
The slope of the line is therefore \frac{b}{ a}. So, the equation of the line can be written in point-slope form, using the point ( a, b) and the slope, as
\( y-b = \frac{b}{a }(x-a)\).
Expanding this equation gives, \(y-b = (\frac{b}{a})x -b\) or just \(y = \frac{b}{a}x\).
Plug y = 0 to find the x-intercept, giving an x-intercept of 0. The statement is sufficient.
STATEMENT (2) SUFFICIENT: There are two cases to consider:
First, we might have m = n = 0. In this case, the point (0, 0) is on line L, so that point is the x-intercept.
Second, we might have any other numbers m and n. In this case, ( m, n) and (– m, – n) are two different points, so their midpoint (which is distinct from either of them) must also be on the line.
The midpoint of ( m, n) and (– m, – n) is (0, 0), so the x-intercept of the line must be 0.
You can also solve the second case by solving for the equation of the line.
If the line passes through two different points ( m, n) and (– m, – n), then standard algebra techniques yield the equation \(y = \frac{m}{n} x\) for the line.
Plug y = 0 to find the x-intercept, revealing an x-intercept of (0, 0).
In each case the x-intercept is 0, so the statement is sufficient.